cos関数の合成

２つのcos関数 $3$\displaystyle{\hspace{1}{r}_{1}\cos \hspace{1}{{\theta}}_{1}}$ , $3$\displaystyle{\hspace{1}{r}_{2}\cos \hspace{1}{{\theta}}_{2}}$ の合成

$3$\displaystyle{ \hspace{1}{r}_{1}\cos \hspace{1}{{\theta}}_{1}+\hspace{1}{r}_{2}\cos \hspace{1}{{\theta}}_{2}=r\cos ${\theta}+{\varphi}$ }$

ここで，右辺の $3$\displaystyle{r}$ , $3$\displaystyle{{\theta}}$ , $3$\displaystyle{{\varphi}}$ $3$\displaystyle{ $- {\pi}< {\varphi}\leq {\pi}$ }$ は次式で与えられる．

$3$\displaystyle{ r=\sqrt{ r_{1}^{2}+r_{2}^{2}+2\hspace{1}{r}_{1}\hspace{1}{r}_{2}\cos $\hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2}$ } }$

$3$\displaystyle{ {\theta}=\frac{\hspace{2} \hspace{1}{{\theta}}_{1}+\hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$

$3$\displaystyle{ \tan {\varphi}=\frac{\hspace{2} \hspace{1}{r}_{1}- \hspace{1}{r}_{2} \hspace{2}}{\hspace{2} \hspace{1}{r}_{1}+\hspace{1}{r}_{2} \hspace{2}}\tan \frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$

ただし， $3$\displaystyle{\hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2}=\pm {\pi}}$ のとき， $3$\displaystyle{r=\hspace{1}{r}_{1}- \hspace{1}{r}_{2}}$ ， $3$\displaystyle{{\varphi}=\pm {\pi}/2 }$ とする．

または，次式のようにとっても構わない．

$3$\displaystyle{ r=\sqrt{ r_{1}^{2}+r_{2}^{2}+2\hspace{1}{r}_{1}\hspace{1}{r}_{2}\cos $\hspace{1}{{\theta}}_{1}+\hspace{1}{{\theta}}_{2}$ } }$

$3$\displaystyle{ {\theta}=\frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$

$3$\displaystyle{ \tan {\varphi}=\frac{\hspace{2} \hspace{1}{r}_{1}- \hspace{1}{r}_{2} \hspace{2}}{\hspace{2} \hspace{1}{r}_{1}+\hspace{1}{r}_{2} \hspace{2}}\tan \frac{\hspace{2} \hspace{1}{{\theta}}_{1}+\hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$

ただし， $3$\displaystyle{\hspace{1}{{\theta}}_{1}+\hspace{1}{{\theta}}_{2}=\pm {\pi}}$ のとき， $3$\displaystyle{r=\hspace{1}{r}_{1}- \hspace{1}{r}_{2}}$ ， $3$\displaystyle{{\varphi}=\pm {\pi}/2 }$ とする．

■ 導出

⇒ 複素数を用いた導出（複素平面での幾何学的な意味）

$3$\displaystyle{ \hspace{1}{r}_{1}\cos \hspace{1}{{\theta}}_{1}+\hspace{1}{r}_{2}\cos \hspace{1}{{\theta}}_{2} }$

$3$\displaystyle{ =\frac{\hspace{2} \hspace{1}{r}_{1}+\hspace{1}{r}_{2} \hspace{2}}{\hspace{2}2\hspace{2}}\left({ \cos \hspace{1}{{\theta}}_{1}+\cos \hspace{1}{{\theta}}_{2} }\right) }$ $3$\displaystyle{ +\frac{\hspace{2} \hspace{1}{r}_{1}- \hspace{1}{r}_{2} \hspace{2}}{\hspace{2}2\hspace{2}}\left({ \cos \hspace{1}{{\theta}}_{1}- \cos \hspace{1}{{\theta}}_{2} }\right) }$

$3$\displaystyle{ =$\hspace{1}{r}_{1}+\hspace{1}{r}_{2}$\cos \frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}}\cos \frac{\hspace{2} \hspace{1}{{\theta}}_{1}+\hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$ $3$\displaystyle{ - $\hspace{1}{r}_{1}- \hspace{1}{r}_{2}$\sin \frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}}\sin \frac{\hspace{2} \hspace{1}{{\theta}}_{1}+\hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$ （ $3$\displaystyle{{}^\bullet\limits{ }_\bullet {}^\bullet }$ 和積の公式）

上式において

$3$\displaystyle{ a=$\hspace{1}{r}_{1}+\hspace{1}{r}_{2}$\cos \frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$ , $3$\displaystyle{ b=$\hspace{1}{r}_{1}- \hspace{1}{r}_{2}$\sin \frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$ , $3$\displaystyle{ {\theta}=\frac{\hspace{2} \hspace{1}{{\theta}}_{1}+\hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$

とおくと

$3$\displaystyle{ \hspace{1}{r}_{1}\cos \hspace{1}{{\theta}}_{1}+\hspace{1}{r}_{2}\cos \hspace{1}{{\theta}}_{2}=a\cos {\theta}- b\sin {\theta} }$

$3$\displaystyle{ =\sqrt{ {a}^{2}+{b}^{2} }$\frac{\hspace{2}a\hspace{2}}{\hspace{2} \sqrt{ {a}^{2}+{b}^{2} } \hspace{2}}\cos {\theta}- \frac{\hspace{2}b\hspace{2}}{\hspace{2} \sqrt{ {a}^{2}+{b}^{2} } \hspace{2}}\sin {\theta}$ }$

$3$\displaystyle{ =r$\cos {\theta}\cos {\varphi}- \sin {\theta}\sin {\varphi}$ }$ $3$\displaystyle{ =r\cos ${\theta}+{\varphi}$ }$ （ $3$\displaystyle{{}^\bullet\limits{ }_\bullet {}^\bullet }$ 加法定理）

が得られる．ここで， $3$\displaystyle{ r=\sqrt{ {a}^{2}+{b}^{2} } }$ , $3$\displaystyle{ \cos {\varphi}=\frac{\hspace{2}a\hspace{2}}{\hspace{2} \sqrt{ {a}^{2}+{b}^{2} } \hspace{2}} }$ , $3$\displaystyle{ \sin {\varphi}=\frac{\hspace{2}b\hspace{2}}{\hspace{2} \sqrt{ {a}^{2}+{b}^{2} } \hspace{2}} }$ とおいた（三角関数の合成）．

また，

$3$\displaystyle{ {a}^{2}+{b}^{2} }$ $3$\displaystyle{ ={ $\hspace{1}{r}_{1}+\hspace{1}{r}_{2}$ }^{2}{ \cos }^{2}\frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$ $3$\displaystyle{ +{ $\hspace{1}{r}_{1}- \hspace{1}{r}_{2}$ }^{2}{ \sin }^{2}\frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$

$3$\displaystyle{ =$ r_{1}^{2}+2\hspace{1}{r}_{1}\hspace{1}{r}_{2}+r_{2}^{2} ${ \cos }^{2}\frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$ $3$\displaystyle{ +$ r_{1}^{2}- 2\hspace{1}{r}_{1}\hspace{1}{r}_{2}+r_{2}^{2} ${ \sin }^{2}\frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$

$3$\displaystyle{ =$ r_{1}^{2}+r_{2}^{2} $$ { \cos }^{2}\frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}}+{ \sin }^{2}\frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} $ }$ $3$\displaystyle{ +2\hspace{1}{r}_{1}\hspace{1}{r}_{2}$ { \cos }^{2}\frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}}- { \sin }^{2}\frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} $ }$

$3$\displaystyle{ =r_{1}^{2}+r_{2}^{2}+2\hspace{1}{r}_{1}\hspace{1}{r}_{2}\cos $\hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2}$ }$ （ $3$\displaystyle{{}^\bullet\limits{ }_\bullet {}^\bullet }$ 2倍角の公式）

より

$3$\displaystyle{ r=\sqrt{ {a}^{2}+{b}^{2} } }$ $3$\displaystyle{ =\sqrt{ r_{1}^{2}+r_{2}^{2}+2\hspace{1}{r}_{1}\hspace{1}{r}_{2}\cos $\hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2}$ } }$

であり， $3$\displaystyle{{\varphi}}$ は

$3$\displaystyle{ \tan {\varphi}=\frac{\hspace{2} \sin {\varphi} \hspace{2}}{\hspace{2} \cos {\varphi} \hspace{2}}=\frac{\hspace{2}b\hspace{2}}{\hspace{2}a\hspace{2}} }$ $3$\displaystyle{ =\frac{\hspace{2} $\hspace{1}{r}_{1}- \hspace{1}{r}_{2}$\sin \frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}}{\hspace{2} $\hspace{1}{r}_{1}+\hspace{1}{r}_{2}$\cos \frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}} }$ $3$\displaystyle{ =\frac{\hspace{2} \hspace{1}{r}_{1}- \hspace{1}{r}_{2} \hspace{2}}{\hspace{2} \hspace{1}{r}_{1}+\hspace{1}{r}_{2} \hspace{2}}\tan \frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$

を満たす角である．（導出完了）

（※）導出の最初の展開式において

$3$\displaystyle{ a=$\hspace{1}{r}_{1}+\hspace{1}{r}_{2}$\cos \frac{\hspace{2} \hspace{1}{{\theta}}_{1}+\hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$ , $3$\displaystyle{ b=$\hspace{1}{r}_{1}- \hspace{1}{r}_{2}$\sin \frac{\hspace{2} \hspace{1}{{\theta}}_{1}+\hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$ , $3$\displaystyle{ {\theta}=\frac{\hspace{2} \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$

とおいて同様に進めると，

$3$\displaystyle{ r=\sqrt{ r_{1}^{2}+r_{2}^{2}+2\hspace{1}{r}_{1}\hspace{1}{r}_{2}\cos $\hspace{1}{{\theta}}_{1}+\hspace{1}{{\theta}}_{2}$ } }$ , $3$\displaystyle{ \tan {\varphi}=\frac{\hspace{2} \hspace{1}{r}_{1}- \hspace{1}{r}_{2} \hspace{2}}{\hspace{2} \hspace{1}{r}_{1}+\hspace{1}{r}_{2} \hspace{2}}\tan \frac{\hspace{2} \hspace{1}{{\theta}}_{1}+\hspace{1}{{\theta}}_{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$

が得られる．

また，sin関数の形で合成

$3$\displaystyle{ \hspace{1}{r}_{1}\cos \hspace{1}{{\theta}}_{1}+\hspace{1}{r}_{2}\cos \hspace{1}{{\theta}}_{2}=r\sin ${\theta}+{\varphi}$ }$

したり，sinの合成 $3$\displaystyle{ \hspace{1}{r}_{1}\sin \hspace{1}{{\theta}}_{1}+\hspace{1}{r}_{2}\sin \hspace{1}{{\theta}}_{2} }$ や，sinとcosの合成 $3$\displaystyle{ \hspace{1}{r}_{1}\sin \hspace{1}{{\theta}}_{1}+\hspace{1}{r}_{2}\cos \hspace{1}{{\theta}}_{2} }$ もできる．

ホーム>>カテゴリー分類>>三角関数>>加法定理>>cos関数の合成

最終更新日：2023年2月28日